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Question-193647

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Question Number 193647 by Rupesh123 last updated on 17/Jun/23
Answered by MM42 last updated on 17/Jun/23
both numbers a , b can not be odd. so a=2 or b=2 if a=2 & b=3 ⇒p=2^3 +7×3^2 =71 ✓ if a=3 & b=2⇒ p=3^2 +7×2^3 =65 × if a=2 & b>3 ; b is primer number ⇒ b=3k+1 ⇒p=2^(3k+1) +7×(3k+1)^2 ≡^3 0 × ⇒b=3k+2 ⇒p=2^(3k+2) +7×(3k+2)^2 ≡^3 0 × similary in other cases the result p is not prime number
$${both}\:{numbers}\:{a}\:,\:{b}\:{can}\:{not}\:{be}\:{odd}. \\ $$$${so}\:{a}=\mathrm{2}\:{or}\:{b}=\mathrm{2} \\ $$$${if}\:\:{a}=\mathrm{2}\:\&\:{b}=\mathrm{3}\:\Rightarrow{p}=\mathrm{2}^{\mathrm{3}} +\mathrm{7}×\mathrm{3}^{\mathrm{2}} =\mathrm{71}\:\checkmark \\ $$$${if}\:\:{a}=\mathrm{3}\:\&\:{b}=\mathrm{2}\Rightarrow\:{p}=\mathrm{3}^{\mathrm{2}} +\mathrm{7}×\mathrm{2}^{\mathrm{3}} =\mathrm{65}\:\:× \\ $$$${if}\:\:\:{a}=\mathrm{2}\:\:\&\:{b}>\mathrm{3}\:;\:{b}\:\:{is}\:\:{primer}\:{number} \\ $$$$\Rightarrow\:{b}=\mathrm{3}{k}+\mathrm{1}\:\Rightarrow{p}=\mathrm{2}^{\mathrm{3}{k}+\mathrm{1}} +\mathrm{7}×\left(\mathrm{3}{k}+\mathrm{1}\right)^{\mathrm{2}} \:\overset{\mathrm{3}} {\equiv}\mathrm{0}\:\:× \\ $$$$\Rightarrow{b}=\mathrm{3}{k}+\mathrm{2}\:\Rightarrow{p}=\mathrm{2}^{\mathrm{3}{k}+\mathrm{2}} +\mathrm{7}×\left(\mathrm{3}{k}+\mathrm{2}\right)^{\mathrm{2}} \overset{\mathrm{3}} {\equiv}\:\mathrm{0}\:× \\ $$$${similary}\:\:{in}\:\:{other}\:\:{cases}\:{the}\:{result} \\ $$$${p}\:\:{is}\:{not}\:{prime}\:{number} \\ $$$$ \\ $$
Answered by AST last updated on 18/Jun/23
a and b cannot both be odd,if so,p would be even ⇒2=a^b +7b^a (a,b>2) which is impossible When b=2,p=a^2 +7(2^a ) a^2 ≡1(mod 3),otherwise a=3(giving p=65(contradiction)) Also 7(2^a )≡−7≡2(mod 3)⇒p≡0(mod 3) ⇒p=3=a^2 +7(2^a )>7 which is impossible ⇒b≠2.. So,we consider a=2 next When a=2 and b is odd p=2^b +7b^2 b^2 ≡1(mod 3) otherwise b=3(giving p=71) Since 71 is a prime,we get (a,b)=(2,3) produces a prime Considering b^2 ≡1(mod 3) gives p=3(impossible) ⇒The largest such prime is 71.
$${a}\:{and}\:{b}\:{cannot}\:{both}\:{be}\:{odd},{if}\:{so},{p}\:{would}\:{be}\:{even} \\ $$$$\Rightarrow\mathrm{2}={a}^{{b}} +\mathrm{7}{b}^{{a}} \left({a},{b}>\mathrm{2}\right)\:\:{which}\:{is}\:{impossible} \\ $$$${When}\:{b}=\mathrm{2},{p}={a}^{\mathrm{2}} +\mathrm{7}\left(\mathrm{2}^{{a}} \right) \\ $$$${a}^{\mathrm{2}} \equiv\mathrm{1}\left({mod}\:\mathrm{3}\right),{otherwise}\:{a}=\mathrm{3}\left({giving}\:{p}=\mathrm{65}\left({contradiction}\right)\right) \\ $$$${Also}\:\mathrm{7}\left(\mathrm{2}^{{a}} \right)\equiv−\mathrm{7}\equiv\mathrm{2}\left({mod}\:\mathrm{3}\right)\Rightarrow{p}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right) \\ $$$$\Rightarrow{p}=\mathrm{3}={a}^{\mathrm{2}} +\mathrm{7}\left(\mathrm{2}^{{a}} \right)>\mathrm{7}\:{which}\:{is}\:{impossible} \\ $$$$\Rightarrow{b}\neq\mathrm{2}..\:{So},{we}\:{consider}\:{a}=\mathrm{2}\:{next} \\ $$$${When}\:{a}=\mathrm{2}\:{and}\:{b}\:{is}\:{odd}\:{p}=\mathrm{2}^{{b}} +\mathrm{7}{b}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} \equiv\mathrm{1}\left({mod}\:\mathrm{3}\right)\:{otherwise}\:{b}=\mathrm{3}\left({giving}\:{p}=\mathrm{71}\right) \\ $$$${Since}\:\mathrm{71}\:{is}\:{a}\:{prime},{we}\:{get}\:\left({a},{b}\right)=\left(\mathrm{2},\mathrm{3}\right)\:{produces} \\ $$$${a}\:{prime} \\ $$$${Considering}\:{b}^{\mathrm{2}} \equiv\mathrm{1}\left({mod}\:\mathrm{3}\right)\:{gives}\:{p}=\mathrm{3}\left({impossible}\right) \\ $$$$\Rightarrow{The}\:{largest}\:{such}\:{prime}\:{is}\:\mathrm{71}. \\ $$

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